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Mohammed's question in facebook

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Here is the question

prove that

\(\displaystyle \sinh \left( \tanh^{-1} (x) \right) = \frac{x}{ \sqrt{1-x^2}}\)
 
Last edited:

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Hey Mohammed , We know that :


\(\displaystyle \sinh(x) = \frac{e^{x}-e^{-x}}{2}\)

\(\displaystyle \sinh\left( \tanh^{-1} (x) \right) = \frac{e^{\tanh^{-1} (x)}-e^{-\tanh^{-1} (x)}}{2}\)

We know also :

\(\displaystyle \tanh^{-1} (x) = \ln \left( \sqrt{ \frac{1+x}{1-x}} \right)\)

\(\displaystyle \sinh \left( \tanh^{-1} (x) \right) = \frac{ \sqrt{ \frac{1+x}{1-x} }- \sqrt{ \frac{1-x}{1+x} } }{2}\)

\(\displaystyle \sinh \left( \tanh^{-1} (x) \right) = \frac{x}{ \sqrt{1-x^2}}\)

if you have further questions you can post them in the calculus section .